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## Main Question or Discussion Point

Hi! Just a couple questions on the Compton scattering calculation in P&S. I feel like I'm missing something very simple here but can't quite figure out what it is. On page 166, the amplitude to be evaluated is

$$ i\mathcal M = -ie^2 \epsilon_\mu(k)\epsilon^*_\nu(k^\prime) u_R^\dagger(p^\prime) \sigma^\mu \frac{\overline \sigma \cdot (p-k^\prime)}{-(\omega^2\chi^2+m^2)}\sigma^\nu u_R(p),$$ where ## u_R(p) = \sqrt{2E}(0, 1)^T## and ##u_R(p^\prime) = \sqrt{2E}(1, 0)^T.## P&S argue that the final photon should be right handed otherwise the amplitude vanishes but I don't quite get that. I presumed that if the final photon is right handed then ##\epsilon^\nu = 1/\sqrt{2}(0,1,i,0)^T##, so ##\epsilon_\nu = 1/\sqrt{2}(0,-1,-i,0)^T## and ##\epsilon_\nu^* = (0, -1, i, 0)^T##, then we would have

$$

\sigma^\nu \epsilon_\nu^* = \frac{1}{\sqrt{2}} \bigg[\begin{pmatrix}

0 & 1\\

1 & 0\end{pmatrix} (-1) + \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix} i\bigg] = \begin{pmatrix} 0 & 0 \\ -\sqrt{2} & 0 \end{pmatrix},

$$

But then ##\sigma^\nu \epsilon_\nu^*u_R(p) = \begin{pmatrix} 0 & 0 \\ -\sqrt{2} & 0 \end{pmatrix} \begin{pmatrix} 0 \\ 1 \end{pmatrix} = 0##. What did I do wrong?

Also with Figure 5.6 on the next page, I have pretty much the same question as this guy in StackExchange https://physics.stackexchange.com/questions/390121/spin-vs-helicity-conservation, unfortunately no one gave him an answer yet. My understanding is that, before the collision, we have spin/helicity of the photon = 1, spin of electron -1/2; and after the collision we have helicity of the photon = -1, and spin of electron 1/2, therefore one unit of spin is exchanged, is that correct? Any help is appreciated. Thanks!

$$ i\mathcal M = -ie^2 \epsilon_\mu(k)\epsilon^*_\nu(k^\prime) u_R^\dagger(p^\prime) \sigma^\mu \frac{\overline \sigma \cdot (p-k^\prime)}{-(\omega^2\chi^2+m^2)}\sigma^\nu u_R(p),$$ where ## u_R(p) = \sqrt{2E}(0, 1)^T## and ##u_R(p^\prime) = \sqrt{2E}(1, 0)^T.## P&S argue that the final photon should be right handed otherwise the amplitude vanishes but I don't quite get that. I presumed that if the final photon is right handed then ##\epsilon^\nu = 1/\sqrt{2}(0,1,i,0)^T##, so ##\epsilon_\nu = 1/\sqrt{2}(0,-1,-i,0)^T## and ##\epsilon_\nu^* = (0, -1, i, 0)^T##, then we would have

$$

\sigma^\nu \epsilon_\nu^* = \frac{1}{\sqrt{2}} \bigg[\begin{pmatrix}

0 & 1\\

1 & 0\end{pmatrix} (-1) + \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix} i\bigg] = \begin{pmatrix} 0 & 0 \\ -\sqrt{2} & 0 \end{pmatrix},

$$

But then ##\sigma^\nu \epsilon_\nu^*u_R(p) = \begin{pmatrix} 0 & 0 \\ -\sqrt{2} & 0 \end{pmatrix} \begin{pmatrix} 0 \\ 1 \end{pmatrix} = 0##. What did I do wrong?

Also with Figure 5.6 on the next page, I have pretty much the same question as this guy in StackExchange https://physics.stackexchange.com/questions/390121/spin-vs-helicity-conservation, unfortunately no one gave him an answer yet. My understanding is that, before the collision, we have spin/helicity of the photon = 1, spin of electron -1/2; and after the collision we have helicity of the photon = -1, and spin of electron 1/2, therefore one unit of spin is exchanged, is that correct? Any help is appreciated. Thanks!